definition - civilittee · types of foundations deep foundations: this type is needed when bearing...

Definition

Foundations ( Footings) : The structural memberdischarges the load of the building into the ground

that

This structural member

is constructed under the

surface of the ground.

Slab Column Foundations GroundBeam

Types of Foundations

Shallow

StrapContinuousRaft (mat)CombinedSingle

Deep


Deep Foundations: This type is needed when bearing capacity of shallow soilformation is poor and structure loads are heavy.

A deep foundation is used to transfer a load from a structure through an upper weak layer of soilto a stronger deeper layer of soil.

In addition to supporting structures, piles are also used to anchor structures against uplift forcesand to assist structures in resisting lateral and overturning forces.

This type gains its strength in supporting heavy loads from frictional forces developed between thesurrounding soil and the surface area of the foundation

Example: Piles , Piers


Shallow Foundations: Most common type offoundation used when firm soil strata is not toodeep and soil engineering properties are suitableregardless of the load transferred by the foundationsto the ground.

Bearing Capacity: Strength of soil (load/area)

Example: kg/cm² , t/m² , N/mm² , Ib/ft², etc…

Types of Shallow Foundations

1. Single Footing: it is an isolated footing thatsupports one column.

Shapes: Square, rectangle or circular; it depends onthe cross section of column and space available.


1. Single Footing:


2. Combined Footing: it is a footing that supportsmore than one column, provided all columns sharethe same axis.

Why is it used?

1. Space between columns is short.

2. Loads on columns are high and bearing capacity of soil is low → causes over lapping


2. Combined Footing:

when the two column are so close to each other that their single footings

would overlap.

Distance between footings is very small

When one column is close to a property line or sewer pipe, the Centre of gravity of the

column will not coincide with footing. In such cases, it is necessary to provide combined this

footing with that of the adjacent internal column ( p1>p2) .

P1>P2


3. Wall Footing: this strip footing is designedmainly to supports a wall, but any columns on thewall axis will share the same footing.


3. Wall Footing:


4. Mat (Raft) Footing:

Raft foundation: is a thick concrete slab reinforced withsteel which covers the entire contact area of thestructure like a thick floor.



Why is it used?

1. Mat foundations become a must when structure loads arehigh, bearing capacity of soil is low and soil highlycompressible.

2. Total area of individual footing ΣAf > 60% of plan area ofstructure; it is more economical and safer to choose matfoundation

Chosen the Most Economical

Safe Foundation Depends on

Major properties of soil

Bearing capacity; Strength of soil (load/area); kg/cm² , t/m² , N/mm² , Ib/ft², etc…

Compressibility

Expansive clay

Load on Structure

Space between axis of columns

FOUNDATIONS

,0

0

A

AP

,0

A

P

footing

footing

footing

footing

A

Pq

CB

PA

CBA

P

A

WOP

*1.1

.

*1.1

.*1.1

.

0.1*P

Find the depth :

Shear

Wide shear Punching shear

'2 call fwide '4 call fpunching

Choose the maximum “d”

Ld

dCB

Lq

resistingarea

areashadedq

resistingarea

shearforce

wideall*

)22

(***)(

Wide shear :

Shaded area

Resisting area

q

OR

Bd

dCL

Bq

resistingarea

areashadedq

resistingarea

shearforce

wideall*

)22

(***)(

q

Punching shear

ddCdC

dCdCBLq

resistingArea

V

Punching

Punching

*)](2)(2[

)])((*[*

21

21

Example :

P= 218 t ,

Dimensions of column (50*50)cm

B.C = 1.5 kg/

Find Area of footing ?

2cm

2

2

/130

/65

cmtpunching

cmtwide

all

all

Solution:

1598.144*4

218*1.1

416

*

216

298.15

15

218*1.1

*1.1.

2/15

2/5.1.

q

mBL

LBA

mmA

A

PCB

mtcmkgCB

resistingarea

shearforce

wideall

)(

Shaded area

qShaded area

Resisting area

cmd

d

d

resistingarea

areashadedq

resistingarea

shearforce

wideall

32

654*

)25.02(*4*98.14*)(

dPunching

455.0

*)]5.0(2)5.0(2[

)]5.0)(5.0(4*4[*98.14130

*)]5.0(2)5.0(2[

)]5.0)(5.0(4*4[*98.14

d

ddd

dd

ddd

dd

resistingArea

V

Punching

Punching

Moment (y-axis)

Maximum moment on the face of the column

F

F = q * 4 * 1.75

M= q * 4 * 1.75 * 1.75/2

M= 14.98 * 4*1.75 * 1.75/2

M= 91.7525 t.m

q*4

Example 2 :

column = 30 x 60 cm

Load = 240 KN

B.C =20 t/m

Fv (WB) = 15 t/m

Fv (ps) = 30 t/m

1. Find Ad

A = 1.1 P = 1.1* 240 = 13.2 m2

B.C 20

choose the shape of footing :

we cannot use square footing since the dimensions are controlled by the property

lines.

A = =3.63m x

There fore , we tend to choose rectangular footing

1.5 * 2 =3m

3 * 2 =6m so lets take B= 3m

L= A =4.4m

B

4.4 =2.2 < 3m … OK.

2

4.4m

2. q = 1.1 P = 1.1* 240 = 20 t/m2

Ad 13.2

A. Wide beam shear :

1- short direction :

shaded area = (3)(2.2-0.3-d)= (5.7-3d) m

shear force =(5.7 -3d) * 20

= (114 -60d) ton

Fv = shear force = 15 = 114 -60d d =1.1 m

resisting area 3d

2- long direction :

shaded area = (4.4)(1.5-0.15-d)= (5.94- 4.4d ) m

shear force = (5.94- 4.4d ) *20

= (118.8- 88d) ton

Fv = 118.8- 88d =15 d= 0.77 m

44d

3m

1

2

3- Punching shear

Shaded area = (4.4*3) – ( (0.6+d)(0.3+d) )

= (13.02 - 0.9d -d ) m

Shear force = (13.02 -0.9d - d ) *20

= (260.4 -18d -20d ) ton

Fv = 260.4 – 18d – 20d = 30 d= 1.13 m

1.2d + 4d

… punching shear controls d = 1.13 m

d = 1.15m

Shaded area

= 13.02 -0.9d –d

Resisting area

= 2 * (d)(0.3+ d) + 2*(d)(0.6+d)

=1.8d + 4d

3

4- maximum bending moments

1. M y-y = (3*1.9* 1.9 ) *20 = 108.3 ton-m

2

2. M x-x =(4.4* 1.35 * 1.35 ) *20 = 80.2 ton-m

2

y

1.35=3 – 0.15

2

which has a rectangular shape and carry two columns of unequal loads

p1 p2

𝐴𝑟𝑒𝑎 =1.1 σ 𝑃

𝐵.𝐶

After we find the Area of the footing ,,, we find the dimensions (L&B)

to be 2 < (L/B) < 3

Equivalent of loads should be in the centroid of the footing

Equivalent always be closer to the biggest load

𝜏all=𝑉

𝐴 𝑟𝑒𝑠𝑖𝑠𝑡𝑖𝑛𝑔

from this equation and from Triangle similarity we find out the value of the depth

** it’s not necessary to study the punching shear in combined footing because wide shear controls the depth of footing

v

d

d

Bd

𝑙 −𝑐

2− 𝑑

𝑙

𝑑

2

𝑑

2

ൗ𝑑 2

ൗ𝑑 2

c+d

d d+c

B’ B’

Bc1c1

c2c2c2

•𝜇short = 1.1×𝑃𝐵×𝐵′

× 𝐵′ × (𝐵

2−

𝑐1

2) ×

(𝐵

2−

𝑐1

2)

2

B’= 1.5d+c2

Example : C1= 40 x 40 cm DL =700 KNLL = 700 KN Total service load = 1400 KN.

C2 =50 X50 cm DL =900 KN

LL =1200 KN Total service load =2100 KN

Space between columns = 3m (center to center)

B.C = 220 KN/m2

Fv(wb)= 600 KN/m2 , fv(p.s) =1200 KN/m2

Sol. :

1. A = 1.1* P = 1.1(1400+2100) = 17.5 m2

B.C 220

1.

2. R = P = 3500 KN

3. Find location of R :

MR = 0 = 1400(3-X)- 2100X

X =1.2m

4. Choose L :

1. Add 0.6 to the right.

2. Add 0.2 to the left to cover the column

3. Add 0.2 also to the right to maintain the C.G

4. Add 1m for both sides to put a distance from the face of C1 in order to resist punching.

finally L = 6m.

5. Find B :

B = A = 17.5 = 2.92 m B=3m

L 6

check : B = 3 = 1 OK.

L 6 2

6. Design Area : Ad = 6x3 = 18m2

7. Soil pressure = q = 1.1 p = 214 KN/m2

18

q = 1.1 p = 641.4 KN/m

L

OR q = q x B

8. Draw shear and B.M diagrams :

find ehe load of zero shear :

x from trigonometry "تشابه المثلثات"

769.96 = 1155.14

x 3-x

x=1.2 m.

Check for wide beam shear :

Find Vmax @ d :

1155.14 = Vmax @d .

1.8 1.8-(d-0.25)

Vmax @d = 994.939-641.89d

Resisting area = B*d = 3d

Fv =994.939- 641.89d =600

3d

d= 0.41m take d=0.45m.

1.1*2100 KN1.1*400 KN

1.8m3m1.2m

x

770.4

Vmax@d

= 1155.14

d

1154.86

769.96

q =6417 KN/m

462.024

1039.67 B.M= 1039.67 KN.mLong direction

Check for punching shear :

1155.14 = V@ d/2 1.8 V@d/2 = 2079.252- 577.57d -288.785

1.8 1.8-(d/2 +0.25)

V @d/2 = 994.704 – 320.872 d

Resisting area = (d+0.5)*d *4 = (d2 + 0.5d) *4

Fv = 994.704 – 320.872d = 1200 d = 0.253 m

4(d2 + 0.5d) d= 0.3m

d of footing = 0.45 from W.B shear.

Find the moment in the short direction .

M = P. Arm

P = (1.5 -0.25)* (1.175)*q = 962.5

q = (1.1)(1200) = 655.32 KN/m2

3 * 1.175

Arm = 1.25 = 0.625 m

2

M = 601.6 KN-m

CB

PB

mL

CB

PBL

CB

P

footingA

A

PCB

.

*1.1

1

.

*1.1*

.

*1.1

*1.1.

d – Wide beam shear

L=1m

BWallwt

dtB w 22

d

1*

)22

(*1*

*

)(

d

dtB

q

resistingarea

areashadedq

resistingarea

shearforce

wide

w

all

Moment:

In short direction :

In long direction :

We use As minimum

2

22*)22

(*1*

2

22*

W

W

W

b

tBtB

qM

tB

fM

L=1m

BWallwt

22

wtB

bf

2

22wtB

Example :

load on wall = 20 t/m

B.C = 20 kg/

Find Area of footing and dimensions?

2cm

mtw 2.0

2/8)( mtwideall

m

CB

PB

1.120

20*1.1

.

*1.1

L=1m

1.1mWallm2.0

d2

2.0

2

1.1

d

md

d

d

resistingarea

areashadedq

BL

Wq

all

32.0

1*

)2

2.0

2

1.1(*1*20

8

*

201.1*1

20*1.1

*

*1.1

M (short) :

mtM

M

tBtB

qM

tB

fM

W

W

W

b

.025.2

2

2

2.0

2

1.1

*)2

2.0

2

1.1(*20

2

22*)22

(*1*

2

22*

which be used to carry two columns of unequal loads when distance outside

the column of the heaviest load is limited.

SL

s

footing

BL

AB

L

C

L

AB

CLD

CB

RA

PPR

2

)13

(2

3'

.

*1.1

21

Find X MC1 C2

P1P2

LBsB

L

X

R

D’

D

C

Example :

P1= 100t , Dimensions of column (60*60)cm

P2=80t , Dimensions of column (50*50)cm

B.C = 17.6 kg/

Find Area of footing and dimensions?

2cm

2

21

25.116.17

180*1.1

.

*1.1

13.233.18.0

33.1

3*80*180

18080100

mCB

RA

mC

mX

X

tPPR

footing

80 t 100 t 180 t

1.332.13

Property

line

39.655.3

13.2*32

5.0

2

6.03

3'

L

L

CLD

Take : L=5m

m

BL

AB

m

L

C

L

AB

SL

s

25.325.15

25.11*2

2

25.1)15

39.6(

5

25.11*2

)13

(2

Not contact with soil.

Massive (rigid).

I (strap ) > I (footing).

⁘ σ = 𝑃

𝐴1 ±

6𝑒

𝐿

⁘ The building is safe if: - σ < 𝑩𝒆𝒂𝒓𝒊𝒏𝒈 𝒄𝒂𝒑𝒂𝒄𝒊𝒕𝒚- σ ≠ 𝒏𝒆𝒈𝒂𝒕𝒊𝒗𝒆

Eccentricity in foundations is due to either limitation of space available for

footing or to bending moment transmitted by columns especially if we have

frame systems, cause overturning.

P

M = M

e = M

P

M

P

M = M + PX

e = M + PX

P

P

M

X

ex

ey

A. Uniaxial B. Biaxial C. Moment couple

In mechanics :

σ= + P ± MCA I

I = bh^3

12

M = P. X = P. e

Iy = BL^3 , C = L

12 2

σ = P ± (P. e)(L/2) = P ± 6P. eA (BL^3/12) A BL^2

(means compression in the foundation)

σ = P ± (1 ± 6e ) + 0.1PA L A

(in tension zone)

(in compression zone)

x

L/2L/2

Max tension Max comp.

B/2

B/2

Y

Y

P

P

σA = P (1+ 6e ) + 0.1P < B.C

A L A

σB = P (1- 6e ) + 0.1P < B.C and +ve

A L A

SAFE

But if σ = -ve the stress will be tensile so the control between the RC

footing and soil is lost ! (fail)

To ensure that there is no tensile stresses

e must be within L from y axis.

6

6e < 1 e= L (max limit)

L 6

UNIAXIAL ECCENTRICITY CASES:

Example :

For 4 x 2 single footing that supports a column with 80 ton load, B.C = 15 t/m2 . Check the safety of

footing against eccentricity.

M = (80)(0.4) = 32 t.m

e = 0.4m

σA = P (1+ 6e ) + 0.1P

A L A

= 80 (1 + 6(0.4) ) + 0.1*80 = 17 t/m2 > 15 t/m2 not safe

8 4 8

σB = P (1- 6e ) + 0.1P = 5 t/m2 < B.C and +ve safe.

A L A

0.4m

EX : for the shown single footing 3 x 4 m , p= 150 ton , B.C = 15 t/m2

A. check the safety against eccentricity.

B. Draw soil pressure on the footing.

A. e = 1m = M

P

σ = P (1+ 6e ) + 0.1P

A L A

= 32.5 t/m2 > B.C (not safe)

σ = P (1- 6e ) + 0.1P

A L A

= -5 t/m2 < BC but –ve (not safe)

1.5m

1.5m

1m1m

-5*3 = -15 t/m

3*32.5 = 97.5 t/m

Soil pressure

P

B.

definition - civilittee · types of foundations deep foundations: this type is needed when bearing...

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