Foundation Examples Dr, Basma.pdf

Foundation Engineering 2Design of Mat (Raft) Footings

Dr. Adnan A. Basma

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DESIGN OF MAT (RAFT) FOUNDATIONSDesign Steps and Equations

For an example on Design of Mat Foundations click here

3D View of Mat (Raft) Foundation

x

y

Pu1

Pu5Pu4

Pu2 Pu3

Pu8Pu7

Pu6

Pu9

A

IHG

FED

CB

Pu

ey

ex

B

Top View of Mat (Raft) Foundation

L

L1 L2

B1

B2

Column dimensions l i x bi


Dr. Adnan A. Basma

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STEP 1 – CHECK SOIL PRESSURE FOR SELECTED DIMENTION

Allowable load P = !Pi

Ultimate load Piu = ! [1.4DLi + 1.7LLi]

Ultimate ratio ru =P

Pu , Ultimate pressure qu = qa x ru

Locate the resultant load Pu

In x- direction: !My-axis = 0, " # " #

u

17u1u29u3ux P

LPPLPPe

$%$&

In y- direction: !Mx-axis = 0, " # " #

u

29u7u13u1uy P

BPPBPPe

$%$&

Applied ultimate pressure, qu,applied = ''

(

)

**

+

,--

x

x

y

yu

I

yM

I

xM

A

P

Where A = Area = BL

Mx = Pu ey and My = Pu ex

Ix = ! B L3 and Iy = ! L B3

For the mat shown in Top View, the following sign convention

is used to estimate qu,applied

x (%) x (+)

A B C

y (+)

D

G

E

H

F

I

y (%)


Dr. Adnan A. Basma

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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 whereB' = B/4.

x

y

A

IHG

FED

CB

B

L

L1 L2

B1

B2

J K

L M

B'

B'

B'

B'

I

III

II


Dr. Adnan A. Basma

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Calculations for Strip ABC:

a) The average uniform soil reaction, quI = 2

qq uCuA $

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) = 2

PQ ABC,u$

e) Load multiplying factor FABC = ABC,u

)ABC(avg,u

P

P

f) The modified loads on this strip P'ui = (FABC) x (Pui)

g) Modified Average soil pressure qu, modified = quI x ''(

)

**+

,

Q

P )ABC(avg,u

h) The pressure distribution along the length of the strip

qu, L (ABC) = L

'P ui0

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

B') where:

quII = 2

qq uFuD $

quIII = 2

qq uuG I$

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.

P'u1 P'u2 P'u3

AB

C

L1 L2

%

Top Steel betweenColumn 1 and 2


qu, L (ABC)

% M(kN.m)

+ V(kN)


Dr. Adnan A. Basma

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+

Moment drawn on tension side

Bottom Steelunder Column 2


Dr. Adnan A. Basma

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

A B C

D

G

E

H

F

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 momentdiagram) 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 toEquations for Reinforcement or the percent reinforcement can beobtained directly from Percent Steel Tables).

For an example on Design of Mat Footings click here

I II III

Foundation Engineering 2Example on Design of Mat Foundations

Dr. Adnan A. Basma

1 of 9

FOUNDATION ENGINEERING 2Mat Foundations (Design Equations)

Design Example

Design the mat foundation for the plan shown below. All column dimensions are 50cm 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

y

0.25 m

A M B N CDL= 200 kNLL= 200 kN

DL = 250 kNLL = 250 kN

DL=250 kNLL=200 kN

M N

D E FDL=800 kNLL= 700 kN

0.44m

DL=800 kNLL= 700 kN

DL=650 kNLL=550 kN

Pu 0.1 mO P x

G H IDL=800 kNLL= 700 kN

DL=800 kNLL= 700 kN

DL=650 kNLL=550 kN

Q R

DL=200 kNLL= 200kN

DL = 250 kNLL = 250 kN

DL=200 kNLL=150 kN

J O K P L

0.25 m 0.25 m

Plan of Mat Foundation with column loads and dimensions

4.25 m 8 m 4.25 m

8 m 8 m

16.5 m

21.5

m

7 m

7 m

7 m

7 m

7 m

3.75

m 3

.75 m


Dr. Adnan A. Basma

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Pu

The table below shows the calculations for the loads:

Column DL, kN LL, kN P, kN Pu, kN

A 200 200 400 620B 250 250 500 775C 250 200 450 690D 800 700 1500 2310E 800 700 1500 2310F 650 550 1200 1845G 800 700 1500 2310H 800 700 1500 2310I 650 550 1200 1845J 200 200 400 620K 250 250 500 775L 200 150 350 535

Total Loads = 11000 16945

ru = 1.54

Ultimate pressure qu = qa x ru = 60 x 1.54 = 92.4 kPa

Location of the resultant load Pu

In x- direction: !My-axis = 0,

" # " #16945

853518451845690862023102310620xe

$%%%&$%%%'

= 0.44 m

In y- direction: !Mx-axis = 0,

" # " # " # " #16945

10.5535775620x3.5184523102310.5184523102310x10.569075620 x-3x7ye

%%&%%%%%%%'

= 0.1 m


Dr. Adnan A. Basma

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Applied ultimate pressure, qu,applied = ((

)

*

++

,

-..

x

x

y

yu

I

yM

I

xM

A

P

Where A = Area = BL = (16.5) x (21.5) = 354.75 m2

Mx = Pu ey = (16945)(0.1) = 1694.5 kN.m

My = Pu ex = (16945)(0.44) = 7455.8 kN.m

Ix = 12

1 B L3 =

12

1 (16.5) (21.5)3 = 13665 m4

Iy = 12

1 L B3 =

12

1 (21.5) (16.5)3 = 8050 m4

Therefore,

qu,applied = (()

*

++,

-..

13665

y5.1694

8050

x8.7455

75.354

16945

= " #y124.0x93.076.47 ..

Point x y Sign for x Sign for y qu,applied, kPa

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

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.


Dr. Adnan A. Basma

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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 = 2

qq 2edgeu1Edge,u %

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 = 2

PQ total,ui %

e) Load multiplying factor F = total,u

avg,u

P

P

f) The modified loads on this strip P'ui = (F) x (Pui)

g) Modified Average soil pressure qu, modified = qu x (()

*

++,

-

i

avg,u

Q

P


qu, L = L

'P ui/


Dr. Adnan A. Basma

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The following table presents the calculations for the selected strips:

Strip Bi, m Point qu,kPa

qu,avgkPa

Q,kN

Total Pu,kN

Pu, avg,kN

qu, modkPa F

A 56.77 AMOJ 4.25

J 54.1055.44 5065.37 5860.00 5462.69 59.47 0.932

B 49.09 MNPO 8.00

K 46.4347.76 8214.72 6170.00 7192.36 40.97 1.166

C 41.42 NCLP 4.25

L 38.7540.09 3662.77 4915.00 4288.88 45.94 0.873

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 P'u, kN qu,L, kN/mAMOJ A 200 200 400 620 577.84F=0.932 D 800 700 1500 2310 2152.92

G 800 700 1500 2310 2152.92J 200 200 400 620 577.84

Total = 2000 1800 3800 5860 5461.52

254.02

MNPO B 250 250 500 775 903.65F=1.166 E 800 700 1500 2310 2693.46

H 800 700 1500 2310 2693.46K 250 250 500 775 903.65

Total = 2100 1900 4000 6170 7194.22

334.61

NCLP C 250 200 450 690 602.37F=0.873 F 650 550 1200 1845 1610.685

I 650 550 1200 1845 1610.685L 200 150 350 535 467.055

Total = 1750 1450 3200 4915 4290.795

199.57

Total Loads = 5850 5150 11000 16945

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

1265.4

-510.8

888.1

510.8

-63.4

63.4

00

0

500

1000

1500

0 5 10 15 20

Strip AMOJ (B' = 4.25 m)

577.8 kN 2152.9 kN 2152.9 kN 577.8 kN

A D G J

7 m 7 m 7 m0.25 m 0.25 m

254.02 kN/m

-5

+ V(kN)

&

2660.5

7.9

1106.3

2660.5

7.90

00

1000

1500

2000

2500

3000

0 5 10 15 20

-1265.4-888 .1

-1500

-1000

Max. moment between columnsSteel at the bottom

5 + M(kN.m)

&


Dr. Adnan A. Basma

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-503.0 -503.0-1000

-500

Moment drawn on compression side

Max. moment under columnsSteel at the top

Strip MNPO (B' = 8.0 m)

903.7 kN 2693.5 kN 2693.5 kN 903.7 kN

B E H K

10.5 449.1

2496.2

10.5

2496.2

0

00

1000

1500

2000

2500

3000

0 5 10 15 20

7 m 7 m 7 m0.25 m 0.25 m

334.6 kN/m

-1524.9

83.6

-83.6

814.7

1169.8

-1169.8

-814.7

1524.9

-2000

-1500

-1000

00

0

00

1000

1500

2000

0 5 10 15 20


5 + M(kN.m)

&

-5

5

+ V(kN)

&


Dr. Adnan A. Basma

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-982.4 -982.4-1500

-1000

-500


-978.8

845.0

-554.4

-766.6

632.6

420.4

-49.7

50.1

-1500

-1000

-500

0

500

1000

0 5 10 15 20

6.3

1023.5

6.3

554.7

0

500

1000

1500

0 5 10 15 20

Strip NCLP (B' = 4.25 m)

602.4 kN 1610.7 kN 1610.7 kN 467.1 kN

C F I L

7 m 7 m 7 m0.25 m 0.25 m

199.6 kN/m


+ M(kN.m)

&

+ V(kN)

&


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-762.3

-446.6

-1842.1

-2000

-1500

-1000

-500




Dr. Adnan A. Basma

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The table below was prepared for the columns and d’ was estimated usingStructural 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

E & H 4 2.00 2310 0.54

F & I 3 1.50 1845 0.53

L 2 1.00 535 0.25


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

Strip B', m Location Mu, kN.m Mu, kN.m/m p, % AS (cm2/m)+ Reinforcement

AMJO 4.25 0-5 m / top 503 118.4 0.11** 33.2 030@25 cm c-c5-17 m / bottom 2660.5 626.0 0.62.. 40.3 030@20 cm c-c17-21.5 m / top 503 118.4 0.11** 33.2 030@25 cm c-c

MNPO 8 0-5 m / top 982.4 122.8 0.11** 33.2 030@25 cm c-c5-16.5 m / bottom 2496.2 312.0 0.30** 33.2 030@25 cm c-c16.5-21.5 m / top 982.4 122.8 0.11** 33.2 030@25 cm c-c

MNPO 4.25 0-6 m / top 762.3 179.4 0.18** 33.2 030@25 cm c-c6-9 m / bottom 1023.5 240.8 0.24** 33.2 030@25 cm c-c9-13 m / top 446.6 105.1 0.10** 33.2 030@25 cm c-c

1314.5 m / bottom 554.7 130.5 0.12** 33.2 030@25 cm c-c14.5-21.5 m / top 1842.1 433.4 0.43** 33.2 030@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-directionmoments (step 2) by considering (refer to page 1) strips: ACNM (B' = 3.75m), MNPO (B' = 7 m), OPRQ (B' = 7m) and QRLJ (B' = 3.75 m).

Maximum

Use d' = 0.65 m

Foundation Engineering 2 / Review Depth of Concrete Footings by Punching Shear

Dr. Adnan A. Basma

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

w w w

l l l 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 0.17 0.15 0.14 0.13 0.12 0.11 0.10 0.09 0.08 0.08 0.07 400 0.22 0.20 0.18 0.16 0.15 0.14 0.13 0.12 0.11 0.10 0.10 500 0.26 0.23 0.21 0.20 0.18 0.17 0.16 0.14 0.14 0.13 0.12 600 0.29 0.27 0.25 0.23 0.21 0.20 0.18 0.17 0.16 0.15 0.14 700 0.33 0.30 0.28 0.26 0.24 0.22 0.21 0.20 0.18 0.17 0.16 800 0.36 0.33 0.31 0.29 0.27 0.25 0.23 0.22 0.21 0.20 0.18 900 0.39 0.36 0.34 0.31 0.29 0.27 0.26 0.24 0.23 0.22 0.21

1000 0.42 0.39 0.37 0.34 0.32 0.30 0.28 0.27 0.25 0.24 0.23 1100 0.45 0.42 0.39 0.37 0.34 0.32 0.30 0.29 0.27 0.26 0.25 1200 0.48 0.45 0.42 0.39 0.37 0.35 0.33 0.31 0.29 0.28 0.26 1300 0.50 0.47 0.44 0.42 0.39 0.37 0.35 0.33 0.31 0.30 0.28 1400 0.53 0.50 0.47 0.44 0.41 0.39 0.37 0.35 0.33 0.32 0.30 1500 0.55 0.52 0.49 0.46 0.44 0.41 0.39 0.37 0.35 0.34 0.32 1600 0.58 0.54 0.51 0.48 0.46 0.43 0.41 0.39 0.37 0.35 0.34 1700 0.60 0.57 0.54 0.51 0.48 0.45 0.43 0.41 0.39 0.37 0.36 1800 0.62 0.59 0.56 0.53 0.50 0.47 0.45 0.43 0.41 0.39 0.37 1900 0.65 0.61 0.58 0.55 0.52 0.49 0.47 0.45 0.43 0.41 0.39 2000 0.67 0.63 0.60 0.57 0.54 0.51 0.49 0.47 0.45 0.43 0.41 2100 0.69 0.65 0.62 0.59 0.56 0.53 0.51 0.48 0.46 0.44 0.42 2200 0.71 0.67 0.64 0.61 0.58 0.55 0.53 0.50 0.48 0.46 0.44 2300 0.73 0.69 0.66 0.63 0.60 0.57 0.54 0.52 0.50 0.48 0.46 2400 0.75 0.71 0.68 0.65 0.62 0.59 0.56 0.54 0.51 0.49 0.47 2500 0.77 0.73 0.70 0.66 0.63 0.61 0.58 0.55 0.53 0.51 0.49 2600 0.79 0.75 0.71 0.68 0.65 0.62 0.60 0.57 0.55 0.53 0.50 2700 0.81 0.77 0.73 0.70 0.67 0.64 0.61 0.59 0.56 0.54 0.52 2800 0.82 0.79 0.75 0.72 0.69 0.66 0.63 0.60 0.58 0.56 0.54 2900 0.84 0.80 0.77 0.74 0.70 0.67 0.65 0.62 0.60 0.57 0.55 3000 0.86 0.82 0.79 0.75 0.72 0.69 0.66 0.64 0.61 0.59 0.57 3100 0.88 0.84 0.80 0.77 0.74 0.71 0.68 0.65 0.63 0.60 0.58 3200 0.89 0.86 0.82 0.79 0.75 0.72 0.69 0.67 0.64 0.62 0.59 3300 0.91 0.87 0.84 0.80 0.77 0.74 0.71 0.68 0.66 0.63 0.61 3400 0.93 0.89 0.85 0.82 0.79 0.75 0.72 0.70 0.67 0.65 0.62 3500 0.94 0.90 0.87 0.83 0.80 0.77 0.74 0.71 0.69 0.66 0.64

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

EXTERIOR COLUMN 3-Way Punch at d'/2 from sides of column

p' = 2l + w

INTERIOR COLUMN 4-Way Punch at d'/2 from sides of column

p' = 2l + 2w

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

p' = l + w


Dr. Adnan A. Basma

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


300 0.16 0.15 0.13 0.12 0.11 0.10 0.09 0.08 0.08 0.07 0.07 400 0.21 0.19 0.17 0.15 0.14 0.13 0.12 0.11 0.10 0.10 0.09 500 0.24 0.22 0.20 0.18 0.17 0.16 0.15 0.14 0.13 0.12 0.11 600 0.28 0.26 0.23 0.22 0.20 0.18 0.17 0.16 0.15 0.14 0.13 700 0.31 0.29 0.26 0.24 0.23 0.21 0.20 0.18 0.17 0.16 0.15 800 0.34 0.32 0.29 0.27 0.25 0.24 0.22 0.21 0.19 0.18 0.17 900 0.37 0.35 0.32 0.30 0.28 0.26 0.24 0.23 0.22 0.20 0.19

1000 0.40 0.37 0.35 0.32 0.30 0.28 0.27 0.25 0.24 0.22 0.21 1100 0.43 0.40 0.37 0.35 0.33 0.31 0.29 0.27 0.26 0.24 0.23 1200 0.46 0.43 0.40 0.37 0.35 0.33 0.31 0.29 0.28 0.26 0.25 1300 0.48 0.45 0.42 0.40 0.37 0.35 0.33 0.31 0.30 0.28 0.27 1400 0.51 0.47 0.44 0.42 0.39 0.37 0.35 0.33 0.32 0.30 0.29 1500 0.53 0.50 0.47 0.44 0.42 0.39 0.37 0.35 0.33 0.32 0.30 1600 0.55 0.52 0.49 0.46 0.44 0.41 0.39 0.37 0.35 0.34 0.32 1700 0.58 0.54 0.51 0.48 0.46 0.43 0.41 0.39 0.37 0.35 0.34 1800 0.60 0.56 0.53 0.50 0.48 0.45 0.43 0.41 0.39 0.37 0.35 1900 0.62 0.58 0.55 0.52 0.50 0.47 0.45 0.43 0.41 0.39 0.37 2000 0.64 0.60 0.57 0.54 0.51 0.49 0.47 0.44 0.42 0.40 0.39 2100 0.66 0.62 0.59 0.56 0.53 0.51 0.48 0.46 0.44 0.42 0.40 2200 0.68 0.64 0.61 0.58 0.55 0.52 0.50 0.48 0.46 0.44 0.42 2300 0.70 0.66 0.63 0.60 0.57 0.54 0.52 0.49 0.47 0.45 0.43 2400 0.72 0.68 0.65 0.62 0.59 0.56 0.53 0.51 0.49 0.47 0.45 2500 0.74 0.70 0.67 0.63 0.60 0.58 0.55 0.53 0.50 0.48 0.46 2600 0.75 0.72 0.68 0.65 0.62 0.59 0.57 0.54 0.52 0.50 0.48 2700 0.77 0.74 0.70 0.67 0.64 0.61 0.58 0.56 0.54 0.51 0.49 2800 0.79 0.75 0.72 0.69 0.66 0.63 0.60 0.57 0.55 0.53 0.51 2900 0.81 0.77 0.74 0.70 0.67 0.64 0.62 0.59 0.57 0.54 0.52 3000 0.82 0.79 0.75 0.72 0.69 0.66 0.63 0.60 0.58 0.56 0.54 3100 0.84 0.80 0.77 0.73 0.70 0.67 0.65 0.62 0.60 0.57 0.55 3200 0.86 0.82 0.78 0.75 0.72 0.69 0.66 0.63 0.61 0.59 0.56 3300 0.87 0.84 0.80 0.77 0.73 0.70 0.68 0.65 0.62 0.60 0.58 3400 0.89 0.85 0.82 0.78 0.75 0.72 0.69 0.66 0.64 0.61 0.59 3500 0.91 0.87 0.83 0.80 0.76 0.73 0.71 0.68 0.65 0.63 0.61



Dr. Adnan A. Basma

3 of 4



300 0.15 0.14 0.12 0.11 0.10 0.09 0.09 0.08 0.07 0.07 0.06 400 0.19 0.17 0.16 0.14 0.13 0.12 0.11 0.10 0.10 0.09 0.09 500 0.23 0.21 0.19 0.17 0.16 0.15 0.14 0.13 0.12 0.11 0.10 600 0.26 0.24 0.22 0.20 0.19 0.17 0.16 0.15 0.14 0.13 0.12 700 0.30 0.27 0.25 0.23 0.21 0.20 0.18 0.17 0.16 0.15 0.14 800 0.33 0.30 0.28 0.26 0.24 0.22 0.21 0.19 0.18 0.17 0.16 900 0.35 0.33 0.30 0.28 0.26 0.24 0.23 0.21 0.20 0.19 0.18

1000 0.38 0.35 0.33 0.30 0.28 0.27 0.25 0.23 0.22 0.21 0.20 1100 0.41 0.38 0.35 0.33 0.31 0.29 0.27 0.25 0.24 0.23 0.22 1200 0.43 0.40 0.38 0.35 0.33 0.31 0.29 0.27 0.26 0.25 0.23 1300 0.46 0.43 0.40 0.37 0.35 0.33 0.31 0.29 0.28 0.26 0.25 1400 0.48 0.45 0.42 0.39 0.37 0.35 0.33 0.31 0.30 0.28 0.27 1500 0.50 0.47 0.44 0.42 0.39 0.37 0.35 0.33 0.31 0.30 0.28 1600 0.53 0.49 0.46 0.44 0.41 0.39 0.37 0.35 0.33 0.31 0.30 1700 0.55 0.51 0.48 0.46 0.43 0.41 0.39 0.37 0.35 0.33 0.32 1800 0.57 0.53 0.50 0.48 0.45 0.43 0.40 0.38 0.36 0.35 0.33 1900 0.59 0.55 0.52 0.49 0.47 0.44 0.42 0.40 0.38 0.36 0.35 2000 0.61 0.57 0.54 0.51 0.49 0.46 0.44 0.42 0.40 0.38 0.36 2100 0.63 0.59 0.56 0.53 0.50 0.48 0.46 0.43 0.41 0.39 0.38 2200 0.65 0.61 0.58 0.55 0.52 0.50 0.47 0.45 0.43 0.41 0.39 2300 0.67 0.63 0.60 0.57 0.54 0.51 0.49 0.47 0.44 0.42 0.41 2400 0.68 0.65 0.61 0.58 0.56 0.53 0.50 0.48 0.46 0.44 0.42 2500 0.70 0.67 0.63 0.60 0.57 0.55 0.52 0.50 0.47 0.45 0.44 2600 0.72 0.68 0.65 0.62 0.59 0.56 0.54 0.51 0.49 0.47 0.45 2700 0.74 0.70 0.67 0.63 0.60 0.58 0.55 0.53 0.50 0.48 0.46 2800 0.75 0.72 0.68 0.65 0.62 0.59 0.57 0.54 0.52 0.50 0.48 2900 0.77 0.73 0.70 0.67 0.64 0.61 0.58 0.56 0.53 0.51 0.49 3000 0.79 0.75 0.71 0.68 0.65 0.62 0.60 0.57 0.55 0.53 0.50 3100 0.80 0.77 0.73 0.70 0.67 0.64 0.61 0.59 0.56 0.54 0.52 3200 0.82 0.78 0.75 0.71 0.68 0.65 0.63 0.60 0.58 0.55 0.53 3300 0.83 0.80 0.76 0.73 0.70 0.67 0.64 0.61 0.59 0.57 0.54 3400 0.85 0.81 0.78 0.74 0.71 0.68 0.65 0.63 0.60 0.58 0.56 3500 0.86 0.83 0.79 0.76 0.73 0.70 0.67 0.64 0.62 0.59 0.57



Dr. Adnan A. Basma

4 of 4



300 0.14 0.12 0.11 0.10 0.09 0.08 0.08 0.07 0.07 0.06 0.06 400 0.18 0.16 0.14 0.13 0.12 0.11 0.10 0.09 0.09 0.08 0.08 500 0.21 0.19 0.17 0.16 0.14 0.13 0.12 0.11 0.11 0.10 0.09 600 0.24 0.22 0.20 0.18 0.17 0.16 0.15 0.14 0.13 0.12 0.11 700 0.27 0.25 0.23 0.21 0.19 0.18 0.17 0.16 0.15 0.14 0.13 800 0.30 0.28 0.25 0.23 0.22 0.20 0.19 0.18 0.16 0.16 0.15 900 0.33 0.30 0.28 0.26 0.24 0.22 0.21 0.19 0.18 0.17 0.16

1000 0.35 0.33 0.30 0.28 0.26 0.24 0.23 0.21 0.20 0.19 0.18 1100 0.38 0.35 0.32 0.30 0.28 0.26 0.25 0.23 0.22 0.21 0.20 1200 0.40 0.37 0.35 0.32 0.30 0.28 0.26 0.25 0.24 0.22 0.21 1300 0.42 0.39 0.37 0.34 0.32 0.30 0.28 0.27 0.25 0.24 0.23 1400 0.45 0.42 0.39 0.36 0.34 0.32 0.30 0.28 0.27 0.25 0.24 1500 0.47 0.44 0.41 0.38 0.36 0.34 0.32 0.30 0.29 0.27 0.26 1600 0.49 0.46 0.43 0.40 0.38 0.36 0.34 0.32 0.30 0.29 0.27 1700 0.51 0.48 0.45 0.42 0.40 0.37 0.35 0.33 0.32 0.30 0.29 1800 0.53 0.50 0.47 0.44 0.41 0.39 0.37 0.35 0.33 0.32 0.30 1900 0.55 0.51 0.48 0.46 0.43 0.41 0.39 0.37 0.35 0.33 0.32 2000 0.57 0.53 0.50 0.47 0.45 0.42 0.40 0.38 0.36 0.35 0.33 2100 0.58 0.55 0.52 0.49 0.46 0.44 0.42 0.40 0.38 0.36 0.34 2200 0.60 0.57 0.54 0.51 0.48 0.46 0.43 0.41 0.39 0.37 0.36 2300 0.62 0.58 0.55 0.52 0.50 0.47 0.45 0.43 0.41 0.39 0.37 2400 0.64 0.60 0.57 0.54 0.51 0.49 0.46 0.44 0.42 0.40 0.38 2500 0.65 0.62 0.59 0.56 0.53 0.50 0.48 0.46 0.43 0.42 0.40 2600 0.67 0.63 0.60 0.57 0.54 0.52 0.49 0.47 0.45 0.43 0.41 2700 0.69 0.65 0.62 0.59 0.56 0.53 0.51 0.48 0.46 0.44 0.42 2800 0.70 0.67 0.63 0.60 0.57 0.55 0.52 0.50 0.48 0.45 0.44 2900 0.72 0.68 0.65 0.62 0.59 0.56 0.53 0.51 0.49 0.47 0.45 3000 0.73 0.70 0.66 0.63 0.60 0.57 0.55 0.52 0.50 0.48 0.46 3100 0.75 0.71 0.68 0.65 0.62 0.59 0.56 0.54 0.51 0.49 0.47 3200 0.76 0.73 0.69 0.66 0.63 0.60 0.58 0.55 0.53 0.51 0.49 3300 0.78 0.74 0.71 0.67 0.64 0.62 0.59 0.56 0.54 0.52 0.50 3400 0.79 0.76 0.72 0.69 0.66 0.63 0.60 0.58 0.55 0.53 0.51 3500 0.81 0.77 0.74 0.70 0.67 0.64 0.62 0.59 0.57 0.54 0.52


Equation (neglecting qu) :2

'f87.8

P22'p

2'p

'd c

u2

+

−+−

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

Depth Factors for bearing capacity equations

Values of dq (inside table)

Df/B

φ° 0.2 0.4 0.6 0.8 1.0 1.2 1.4 1.6 1.8 2.0

0 1.00 1.00 1.00 1.00 1.00 1.00 1.00 1.00 1.00 1.00 1 1.01 1.01 1.02 1.03 1.03 1.03 1.03 1.03 1.04 1.04 2 1.01 1.03 1.04 1.05 1.05 1.06 1.06 1.07 1.07 1.07 3 1.02 1.04 1.06 1.08 1.07 1.08 1.09 1.10 1.10 1.10 4 1.02 1.05 1.07 1.10 1.10 1.11 1.12 1.12 1.13 1.13 5 1.03 1.06 1.09 1.12 1.11 1.13 1.14 1.15 1.16 1.16 6 1.03 1.07 1.10 1.13 1.13 1.15 1.16 1.17 1.18 1.19 7 1.04 1.08 1.11 1.15 1.15 1.17 1.18 1.19 1.20 1.21 8 1.04 1.08 1.12 1.17 1.16 1.18 1.20 1.21 1.22 1.23 9 1.05 1.09 1.14 1.18 1.18 1.20 1.21 1.23 1.24 1.25 10 1.05 1.10 1.14 1.19 1.19 1.21 1.23 1.24 1.26 1.27 11 1.05 1.10 1.15 1.20 1.20 1.22 1.24 1.26 1.27 1.28 12 1.05 1.11 1.16 1.21 1.21 1.23 1.25 1.27 1.28 1.30 13 1.06 1.11 1.17 1.22 1.22 1.24 1.26 1.28 1.30 1.31 14 1.06 1.11 1.17 1.23 1.23 1.25 1.27 1.29 1.30 1.32 15 1.06 1.12 1.18 1.24 1.23 1.26 1.28 1.30 1.31 1.33 16 1.06 1.12 1.18 1.24 1.24 1.26 1.29 1.30 1.32 1.33 17 1.06 1.12 1.18 1.24 1.24 1.27 1.29 1.31 1.33 1.34 18 1.06 1.12 1.19 1.25 1.24 1.27 1.29 1.31 1.33 1.34 19 1.06 1.13 1.19 1.25 1.25 1.27 1.30 1.32 1.33 1.35 20 1.06 1.13 1.19 1.25 1.25 1.28 1.30 1.32 1.34 1.35 21 1.06 1.13 1.19 1.25 1.25 1.28 1.30 1.32 1.34 1.35 22 1.06 1.13 1.19 1.25 1.25 1.28 1.30 1.32 1.34 1.35 23 1.06 1.13 1.19 1.25 1.25 1.28 1.30 1.32 1.34 1.35 24 1.06 1.13 1.19 1.25 1.25 1.27 1.30 1.32 1.33 1.35 25 1.06 1.12 1.19 1.25 1.24 1.27 1.30 1.31 1.33 1.34 26 1.06 1.12 1.18 1.25 1.24 1.27 1.29 1.31 1.33 1.34 27 1.06 1.12 1.18 1.24 1.24 1.27 1.29 1.31 1.32 1.34 28 1.06 1.12 1.18 1.24 1.24 1.26 1.28 1.30 1.32 1.33 29 1.06 1.12 1.18 1.24 1.23 1.26 1.28 1.30 1.31 1.33 30 1.06 1.12 1.17 1.23 1.23 1.25 1.27 1.29 1.31 1.32 31 1.06 1.11 1.17 1.23 1.22 1.25 1.27 1.29 1.30 1.31 32 1.06 1.11 1.17 1.22 1.22 1.24 1.26 1.28 1.29 1.31 33 1.05 1.11 1.16 1.22 1.21 1.24 1.26 1.27 1.29 1.30 34 1.05 1.10 1.16 1.21 1.21 1.23 1.25 1.27 1.28 1.29 35 1.05 1.10 1.15 1.20 1.20 1.22 1.24 1.26 1.27 1.28 36 1.05 1.10 1.15 1.20 1.19 1.22 1.23 1.25 1.26 1.27 37 1.05 1.10 1.14 1.19 1.19 1.21 1.23 1.24 1.25 1.26 38 1.05 1.09 1.14 1.18 1.18 1.20 1.22 1.23 1.25 1.26 39 1.04 1.09 1.13 1.18 1.17 1.19 1.21 1.23 1.24 1.25 40 1.04 1.09 1.13 1.17 1.17 1.19 1.20 1.22 1.23 1.24 41 1.04 1.08 1.12 1.16 1.16 1.18 1.20 1.21 1.22 1.23 42 1.04 1.08 1.12 1.16 1.15 1.17 1.19 1.20 1.21 1.22 43 1.04 1.08 1.11 1.15 1.15 1.17 1.18 1.19 1.20 1.21 44 1.04 1.07 1.11 1.14 1.14 1.16 1.17 1.18 1.19 1.20 45 1.03 1.07 1.10 1.14 1.13 1.15 1.16 1.17 1.18 1.19

dc 1.08 1.16 1.24 1.32 1.31 1.35 1.38 1.40 1.43 1.44 Prepared by Dr. Adnan Basma

Foundation EngineeringEquations for Concrete Depth of Footings

Dr. Adnan A. Basma

1 of 2

STRUCTURAL DEPTH OF CONCRETE FOOTINGS, d' (m)

PUNCHING SHEAR

Punching shear failure occurs at a distance d'/2 from the sides of the column

b b

l l

For equilibrium,

!F = 0 , Pu " Vc " Fs = 0

where: 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.

EXTERIOR COLUMN3-Way Punch at d'/2from sides of column

INTERIOR COLUMN4-Way Punch at d'/2from sides of column

Foundation EngineeringEquations for Concrete Depth of Footings

Dr. Adnan A. Basma

2 of 2

WIDE BEAM FAILURE (Uniform footing width B)

For wide beam shear, failure is assumed at a distance d' from the side of the column.

For equilibrium

!F = 0 , Fs - Vc = 0

where Fs = qua . Lw.1

Vc = #c . d’ . 1

c# = 4.56 'fc (SI units) with fc’ and #c are in kN/m2

Thus 'dq2

'd2Lcua #$$$$

"""""""" !, re-arranging term and solving

d’ = )q(2

q)L(

uac

ua

%%%%""""#

!

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

d'

Lw Wide beam failure at d' from column

L

Bl

b

Foundation Engineering 2 / Review Equations for Reinforcement of Footings

Dr. Adnan A. Basma

1 of 1

ESTIMATION OF REINFORCEMENT FOR CONCRETE DESIGN OF FOOTINGS

To resist tension in concrete we need reinforcements in such a way that

!"#

$%& '()

2

a'dfAM ySu

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 = 1'f85.0

fA

c

ys

+ indicates As / unit length (1m)

Percent steel p = 1'd

As

+ maxmin ppp ,,

wherey

min f

1400p ) or 0.002 (0.2% )

and pmax )10600f(

10600

f

'f5.0

3y

3

y

c

+-+

+. fc’ and fy in kN/m2

Using the estimated value of AS, calculate number of bars / m = n = b

s

A

A

where Ab = Area of bar with diameter db where db = 4

d 2b/

Spacing of bars, s = 1n

cm100

'

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

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

Foundation EngineeringPunching Shear / Example

Dr. Adnan A. Basma

1 of 1

FOUNDATION ENGINEERINGFooting Depth by Punching Shear

Example 3aDetermine the depth of the footing d' by punching shear for the following conditions:

Loads: DL = 1110 kN, LL = 1022 kNMaterials: fc'= 21 MPa, fy = 415 MPaColumn: 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 kPa 1287.4 = 1021x 8.87 = v 3c

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

Foundation EngineeringReinforcement / Example

Dr. Adnan A. Basma

1 of 2

FOUNDATION ENGINEERINGReinforcement

Example 3bDetermine the reinforcement for the following footing conditions:

Footing: L = 4.1 m and B = 2.2m, d' = 0.65mLoads: DL = 1110 kN, LL = 1022 kNMaterials: fc'= 21 MPa, fy = 415 MPaColumn: 45 cm ! 45 cm (at center of footing)Soil: Ultimate soil pressure qu = 365 kPa

SOLUTION:



Ultimate moments in L and B directions

kN.m/m 608.8 = 2

20.45 - 4.1 365.0

= (L)M

2

u

"#

$%&

'

kN.m/m 139.9 = 2

20.45 - 2.2 365.0

= (B)M

2

u

"#

$%&

'

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

calculations in both directions

A 23.24 = )x1]10[0.85x(21x

A )10(415x = a s3

s3

Substituting in ultimate moment versus steel area equation we get

608.8 = 2

A 23.24 - 0.650 A )10(415x 0.9 s

s3 (

)*

+,-

Re-arranging the terms, the final result becomes

0 = 0.000143 + A 0.057 - A ss2

Advanced Foundation EngineeringReinforcement / Example

Dr. Adnan A. Basma

2 of 2

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

1.5% or 1500. = ) 10600x + 10(415x

10600xx

10415x1021x

0.5x = p33

3

3

3

max

larger is whichever0.002 or 0.0033 = 10415x

1400 = p

3min

0.4% or 0.004 = 0.650x1

0.0026 = p

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 = 165.0100

4.0!! = 0.0026 m2/m or 26 cm2/m.

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)



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

Foundation Engineering 2 / Review Example 5 - Reinforcement

Dr. Adnan A. Basma

1 of 2

FOUNDATION ENGINEERINGReinforcement

Example #5 (Reinforcement)Determine the reinforcement for the following footing conditions:

Footing: L = 4.1 m and B = 2.2m, d' = 0.65mLoads: DL = 1110 kN, LL = 1022 kNMaterials: fc'= 21 MPa, fy = 415 MPaColumn: 45 cm ! 45 cm (at center of footing)Soil: Ultimate soil pressure qu = 365 kPa

SOLUTION:



Ultimate moments in L and B directions

kN.m/m 608.8 = 2

20.45 - 4.1 365.0

= (L)M

2

u

"#

$%&

'

kN.m/m 139.9 = 2

20.45 - 2.2 365.0

= (B)M

2

u

"#

$%&

'

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

calculations in both directions

A 23.24 = )x1]10[0.85x(21x

A )10(415x = a s3

s3

Substituting in ultimate moment versus steel area equation we get

608.8 = 2

A 23.24 - 0.650 A )10(415x 0.9 s

s3 (

)*

+,-

Re-arranging the terms, the final result becomes

0 = 0.000143 + A 0.057 - A ss2

Foundation Engineering 2 / Review Example 5 - Reinforcement

Dr. Adnan A. Basma

2 of 2

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

1.5% or 1500. = ) 10600x + 10(415x

10600xx

10415x1021x

0.5x = p33

3

3

3

max

larger is whichever0.002 or 0.0033 = 10415x

1400 = p

3min

0.4% or 0.004 = 0.650x1

0.0026 = p

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 = 165.0100

4.0!! = 0.0026 m2/m or 26 cm2/m.

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.




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


Dr. Adnan A. Basma

1 of 6

DESIGN OF MAT (RAFT) FOUNDATIONSDesign Steps and Equations

For an example on Design of Mat Foundations click here

3D View of Mat (Raft) Foundation

x

y

Pu1

Pu5Pu4

Pu2 Pu3

Pu8Pu7

Pu6

Pu9

A

IHG

FED

CB

Pu

ey

ex

B

Top View of Mat (Raft) Foundation

L

L1 L2

B1

B2

Column dimensions l i x bi


Dr. Adnan A. Basma

2 of 6





Pu , Ultimate pressure qu = qa x ru

Locate the resultant load Pu

In x- direction: !My-axis = 0, " # " #

u

17u1u29u3ux P

LPPLPPe

$%$&

In y- direction: !Mx-axis = 0, " # " #

u

29u7u13u1uy P

BPPBPPe

$%$&

Applied ultimate pressure, qu,applied = ''

(

)

**

+

,--

x

x

y

yu

I

yM

I

xM

A

P

Where A = Area = BL

Mx = Pu ey and My = Pu ex

Ix = ! B L3 and Iy = ! L B3

For the mat shown in Top View, the following sign convention

is used to estimate qu,applied

x (%) x (+)

A B C

y (+)

D

G

E

H

F

I

y (%)


Dr. Adnan A. Basma

3 of 6

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


The mat is divided into several strips in L-direction as shown below whereB' = B/4.

x

y

A

IHG

FED

CB

B

L

L1 L2

B1

B2

J K

L M

B'

B'

B'

B'

I

III

II


Dr. Adnan A. Basma

4 of 6

Calculations for Strip ABC:

a) The average uniform soil reaction, quI = 2

qq uCuA $

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) = 2

PQ ABC,u$

e) Load multiplying factor FABC = ABC,u

)ABC(avg,u

P

P

f) The modified loads on this strip P'ui = (FABC) x (Pui)

g) Modified Average soil pressure qu, modified = quI x ''(

)

**+

,

Q

P )ABC(avg,u


qu, L (ABC) = L

'P ui0

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

B') where:

quII = 2

qq uFuD $

quIII = 2

qq uuG I$

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.

P'u1 P'u2 P'u3

AB

C

L1 L2

%



qu, L (ABC)

% M(kN.m)

+ V(kN)


Dr. Adnan A. Basma

5 of 6

+

Moment drawn on tension side

Bottom Steelunder Column 2


Dr. Adnan A. Basma

6 of 6

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

A B C

D

G

E

H

F

I


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)


The calculations below are repeated for every strip in L and B direction.

a) Select the appropriate moments for each strip (refer to momentdiagram) 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 toEquations for Reinforcement or the percent reinforcement can beobtained directly from Percent Steel Tables).

For an example on Design of Mat Footings click here

I II III


Dr. Adnan A. Basma

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FOUNDATION ENGINEERING 2Mat Foundations (Design Equations)

Design Example

Design the mat foundation for the plan shown below. All column dimensions are 50cm 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

y

0.25 m

A M B N CDL= 200 kNLL= 200 kN

DL = 250 kNLL = 250 kN

DL=250 kNLL=200 kN

M N

D E FDL=800 kNLL= 700 kN

0.44m

DL=800 kNLL= 700 kN

DL=650 kNLL=550 kN

Pu 0.1 mO P x

G H IDL=800 kNLL= 700 kN

DL=800 kNLL= 700 kN

DL=650 kNLL=550 kN

Q R

DL=200 kNLL= 200kN

DL = 250 kNLL = 250 kN

DL=200 kNLL=150 kN

J O K P L

0.25 m 0.25 m

Plan of Mat Foundation with column loads and dimensions

4.25 m 8 m 4.25 m

8 m 8 m

16.5 m

21.5

m

7 m

7 m

7 m

7 m

7 m

3.75

m 3

.75 m


Dr. Adnan A. Basma

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Pu

The table below shows the calculations for the loads:

Column DL, kN LL, kN P, kN Pu, kN

A 200 200 400 620B 250 250 500 775C 250 200 450 690D 800 700 1500 2310E 800 700 1500 2310F 650 550 1200 1845G 800 700 1500 2310H 800 700 1500 2310I 650 550 1200 1845J 200 200 400 620K 250 250 500 775L 200 150 350 535

Total Loads = 11000 16945

ru = 1.54

Ultimate pressure qu = qa x ru = 60 x 1.54 = 92.4 kPa

Location of the resultant load Pu

In x- direction: !My-axis = 0,

" # " #16945

853518451845690862023102310620xe

$%%%&$%%%'

= 0.44 m

In y- direction: !Mx-axis = 0,

" # " # " # " #16945

10.5535775620x3.5184523102310.5184523102310x10.569075620 x-3x7ye

%%&%%%%%%%'

= 0.1 m


Dr. Adnan A. Basma

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Applied ultimate pressure, qu,applied = ((

)

*

++

,

-..

x

x

y

yu

I

yM

I

xM

A

P

Where A = Area = BL = (16.5) x (21.5) = 354.75 m2

Mx = Pu ey = (16945)(0.1) = 1694.5 kN.m

My = Pu ex = (16945)(0.44) = 7455.8 kN.m

Ix = 12

1 B L3 =

12

1 (16.5) (21.5)3 = 13665 m4

Iy = 12

1 L B3 =

12

1 (21.5) (16.5)3 = 8050 m4

Therefore,

qu,applied = (()

*

++,

-..

13665

y5.1694

8050

x8.7455

75.354

16945

= " #y124.0x93.076.47 ..

Point x y Sign for x Sign for y qu,applied, kPa

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

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.


Dr. Adnan A. Basma

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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 = 2

qq 2edgeu1Edge,u %

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 = 2

PQ total,ui %

e) Load multiplying factor F = total,u

avg,u

P

P

f) The modified loads on this strip P'ui = (F) x (Pui)

g) Modified Average soil pressure qu, modified = qu x (()

*

++,

-

i

avg,u

Q

P


qu, L = L

'P ui/


Dr. Adnan A. Basma

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The following table presents the calculations for the selected strips:

Strip Bi, m Point qu,kPa

qu,avgkPa

Q,kN

Total Pu,kN

Pu, avg,kN

qu, modkPa F

A 56.77 AMOJ 4.25

J 54.1055.44 5065.37 5860.00 5462.69 59.47 0.932

B 49.09 MNPO 8.00

K 46.4347.76 8214.72 6170.00 7192.36 40.97 1.166

C 41.42 NCLP 4.25

L 38.7540.09 3662.77 4915.00 4288.88 45.94 0.873

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 P'u, kN qu,L, kN/mAMOJ A 200 200 400 620 577.84F=0.932 D 800 700 1500 2310 2152.92

G 800 700 1500 2310 2152.92J 200 200 400 620 577.84

Total = 2000 1800 3800 5860 5461.52

254.02

MNPO B 250 250 500 775 903.65F=1.166 E 800 700 1500 2310 2693.46

H 800 700 1500 2310 2693.46K 250 250 500 775 903.65

Total = 2100 1900 4000 6170 7194.22

334.61

NCLP C 250 200 450 690 602.37F=0.873 F 650 550 1200 1845 1610.685

I 650 550 1200 1845 1610.685L 200 150 350 535 467.055

Total = 1750 1450 3200 4915 4290.795

199.57

Total Loads = 5850 5150 11000 16945

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

1265.4

-510.8

888.1

510.8

-63.4

63.4

00

0

500

1000

1500

0 5 10 15 20

Strip AMOJ (B' = 4.25 m)

577.8 kN 2152.9 kN 2152.9 kN 577.8 kN

A D G J

7 m 7 m 7 m0.25 m 0.25 m

254.02 kN/m

-5

+ V(kN)

&

2660.5

7.9

1106.3

2660.5

7.90

00

1000

1500

2000

2500

3000

0 5 10 15 20

-1265.4-888 .1

-1500

-1000


5 + M(kN.m)

&


Dr. Adnan A. Basma

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-503.0 -503.0-1000

-500



Strip MNPO (B' = 8.0 m)

903.7 kN 2693.5 kN 2693.5 kN 903.7 kN

B E H K

10.5 449.1

2496.2

10.5

2496.2

0

00

1000

1500

2000

2500

3000

0 5 10 15 20

7 m 7 m 7 m0.25 m 0.25 m

334.6 kN/m

-1524.9

83.6

-83.6

814.7

1169.8

-1169.8

-814.7

1524.9

-2000

-1500

-1000

00

0

00

1000

1500

2000

0 5 10 15 20


5 + M(kN.m)

&

-5

5

+ V(kN)

&


Dr. Adnan A. Basma

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-982.4 -982.4-1500

-1000

-500


-978.8

845.0

-554.4

-766.6

632.6

420.4

-49.7

50.1

-1500

-1000

-500

0

500

1000

0 5 10 15 20

6.3

1023.5

6.3

554.7

0

500

1000

1500

0 5 10 15 20

Strip NCLP (B' = 4.25 m)

602.4 kN 1610.7 kN 1610.7 kN 467.1 kN

C F I L

7 m 7 m 7 m0.25 m 0.25 m

199.6 kN/m


+ M(kN.m)

&

+ V(kN)

&


Dr. Adnan A. Basma

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-762.3

-446.6

-1842.1

-2000

-1500

-1000

-500




Dr. Adnan A. Basma

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The table below was prepared for the columns and d’ was estimated usingStructural 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

E & H 4 2.00 2310 0.54

F & I 3 1.50 1845 0.53

L 2 1.00 535 0.25


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

Strip B', m Location Mu, kN.m Mu, kN.m/m p, % AS (cm2/m)+ Reinforcement

AMJO 4.25 0-5 m / top 503 118.4 0.11** 33.2 030@25 cm c-c5-17 m / bottom 2660.5 626.0 0.62.. 40.3 030@20 cm c-c17-21.5 m / top 503 118.4 0.11** 33.2 030@25 cm c-c

MNPO 8 0-5 m / top 982.4 122.8 0.11** 33.2 030@25 cm c-c5-16.5 m / bottom 2496.2 312.0 0.30** 33.2 030@25 cm c-c16.5-21.5 m / top 982.4 122.8 0.11** 33.2 030@25 cm c-c

MNPO 4.25 0-6 m / top 762.3 179.4 0.18** 33.2 030@25 cm c-c6-9 m / bottom 1023.5 240.8 0.24** 33.2 030@25 cm c-c9-13 m / top 446.6 105.1 0.10** 33.2 030@25 cm c-c

1314.5 m / bottom 554.7 130.5 0.12** 33.2 030@25 cm c-c14.5-21.5 m / top 1842.1 433.4 0.43** 33.2 030@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-directionmoments (step 2) by considering (refer to page 1) strips: ACNM (B' = 3.75m), MNPO (B' = 7 m), OPRQ (B' = 7m) and QRLJ (B' = 3.75 m).

Maximum

Use d' = 0.65 m

Documents

Foundation Examples Dr, Basma.pdf