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Foundation Design Guide - 13


0

- Subsidence stress at the bottom of element layers:

i

oh

glk

In there:

. gl

+ Ratio: L qu /B qu = 4.69/3.39 =1.4

0

+ gl

=24.62T/m2 : is the settlement stress at the bottom of the conventional foundation;


+ k 0i : is the stress coefficient at the center of the rectangle (L qu /B qu ; 2z i /B qu ) looking at table 2.1 we have the subsidence stress at the bottom of the element layers:

Table 3.16: Calculation table of self-stress and settlement stress


STT

h i (m)

z i (m)

2.z i /B m

γ i

(T/ m3 )

σ i bt

(T/ m2 )

k 0i

σ i gl

(T/ m2 )

P i (T/m 2 )

σ i bt / σ i gl

0


-

-


32.94

1,000

24.62



1

0.678

0.678

0.40

1.95

34.26

0.972

23.93

24.28

1.43

2

0.678

1,356

0.80

1.95

35.58

0.848

20.88

22.40

1.70

3

0.678

2,034

1.20

1.95

36.91

0.682

16.79

18.83

2.20

4

0.678

2,712

1.60

1.95

38.23

0.532

13.10

14.94

2.92

5

0.678

3,390

2.00

1.95

39.55

0.414

10.19

11.65

3.88

6

0.678

4,068

2.40

1.95

40.87

0.325

8.00

9.10

5.11

7

0.678

4,746

2.80

1.95

42.19

0.260

6.40

7.20

6.59








Total

108.40


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Foundation Design Guide - 13


e) Check the subsidence stop condition

- Depth z 7 = 4.746m we have:

7

7

bt42.2T / m 25gl5.6.432.0T / m 2


- So stop calculating settlement at depth Z 7 = 4.746m.

f) Determine total settlement and check settlement

- The final (stable) settlement is determined by the following formula:

n

S i ph

E

ii

i 1 oi

S 0.8 .0.678.108.4 = 0.0784

750

S = 7.84m

- Check for subsidence:

S = 7.84 cm S gh = 10cm (Satisfied) In which:

+ h i =0.678m: is the thickness of the i-th element soil layer;


i;

+ pi : is the average subsidence stress of the i-th element layer;

+ i = 0.8 for all soil layers;

+ E 0i = 75 kg/cm 2 = 750T/m 2 : is the total deformation modulus of the soil element layer

- So with the selected foundation and pile size, the allowable settlement conditions are satisfied.

Q tt

N tt

0

0M 0

tt

h m = 2000

3.2.9 Determine the working height of the foundation


45°

H m

a

45°

P 1 =P 2 P 3 P 4 =P 5 P 6 P 7 =P 8

x

L x t = 2600


300

650

1 4 7

B x t = 1900

b c

B m = 1900

3 6 years

650

l c

300

2 5 8



300 650


650


650


650


300

L m = 3200


Figure 3.20: Calculation and puncture test diagram

3.2.9.1 Determine the minimum height of the foundation

- Minimum height of foundation: H min = 2d+a = 2,300+100 = 700

In there:

+ d =300: is the edge of the pile;

+ a = 100: is the pile section attached to the foundation.

- Choose H m = 1.1m ≥ H min

=> H 0 = H m - a = 1.1 - 0.1 = 1.0m

3.2.9.2 Check height H m under puncture condition

- Penetration tower size:


B xt = b c + 2H 0 = 0.4 +2.1.0 = 2.4m > B m => B xt = B m =1.9m L xt = l c + 2H 0 = 0.6 +2.1.0 = 2.6m

- Force causing penetration:


N gxt


n

i

P tt (outside the penetration tower)

i

- We see: The penetration tower covers the center of all piles. Therefore, we do not need to check for penetration.

3.2.10 Steel calculation

3.2.10.1 Select foundation material

- Choose concrete grade 300 with:

+ R b = 130 kG/cm 2 = 1300 T/m 2

+ R bt = 10 kG/cm 2 = 100 T/m 2

- Select AII load-bearing steel with:

+ R s = R sc = 2800 kg/cm 2

3.2.10.2 Calculation of reinforcement according to the side direction L m

- Bending moment in direction L m :

n

M (y l c )P tt

2

L iii 1


M L = 2.62.4.(0.65 0.6 ) 2.64.7.(1.3 0.6 ) 92.61 Tm

2 2

- Steel reinforcement calculated according to L m direction :

A tt M L

92.61.10 5


= 36.75 cm 2

sL 0.9R

s H 0

0.9.2800.100

- Steel structure according to L m direction :


A

ct

sL min

.BH 0

= 0.1%.190.100 = 19.0 cm 2

- Arrange the steel in the lower region according to the L m direction :

A

SL

sL

d max(A tt

; A ct ) = 36.75 cm 2

SL

+ Choose steel 20 with f a = 3.14 cm 2

A d 36.75

+ Number of steel bars: n dL =

sL = 11.7 trees => Choose n dL = 12 trees

n

1

f a 3.14


+ Reinforcement step: @

= B m 100 1900 100 = 164

dL

dL

12 1


=> Select @ dL =160

+ So the steel is arranged in the direction of edge L in the lower region: 20@160

- Arrangement of structural steel in the upper region according to the L m direction :


A

A

t ct

sL SL

= 19.0 cm 2

+ Choose steel 14 with f a = 1.54 cm 2

At 19.0​

+ Number of steel bars: n tL =

sL = 12.3 trees => Choose n tL = 13 trees

f a 1.54


+ Reinforcement step: @

= B m 100 1900 100 = 150

tL

tL

13 1

n

1

+ So the steel is arranged in the direction of edge L in the upper region: 14@150

3.2.10.3 Calculation of reinforcement according to side direction B m

- Bending moment in direction B m :

n

M (x b c )P tt

2

B iii 1


M B = (55.5 60.1 64.7).(0.65 0.4 ) 81.14 Tm

2

- Steel reinforcement calculated according to B m direction :

A tt MB ​

81.14.10 5


= 32.20cm 2

sB 0.9R

s H 0

0.9.2800.100

- Steel structure according to direction B m :


A

ct

sB min

.L m

.H 0

= 0.1%.320.100 = 32 cm 2

- Arrange the steel in the lower region according to the B m direction :

A d max(A tt ; A ct )= 32.20 cm 2

sB sB sB

+ Choose steel 16 with f a = 2.01 cm 2

A d 32.20

+ Number of steel bars: n dB =

sB = 16 trees => Choose n dB = 16 trees

f a 2.01


+ Reinforcement step: @

= L m 100 3200 100 = 207

dB


=> Select @ dB =200

n dB 1 16 1

+ So the steel is arranged in the direction of edge B m in the lower region: 16@200

- Arrangement of structural steel in the upper region according to direction B m :


A

sB

sB

t A ct =32.0 cm 2

+ Choose steel 14 with f a = 1.54 cm 2

A t 32.0

+ Number of steel bars: n tB =

sB = 20.8 trees => Choose n tB = 21 trees

f a 1.54


+ Reinforcement step: @

= L m 100 3200 100 = 155

tB


=>Select @ tB = 150

n tB 1

21 1

+ So the steel is arranged in the direction of edge B m in the upper region: 14@150

3.2.11 Drawing representation

- Foundation drawing: Figure 3.21

- Pile drawing: Figure 3.22




Column steel


NATURAL GROUND


Tripod


4 ị14a150 ị14a150 3


5

2 in 12


Concrete foundation stone 40x60 grade 100


1 ị20a160 ị16a200 2


SECTION AA

TL:1/25


ị14a150 3


AA

ị14a150 4

ị16a200 2

B


ị20a160 1



* STONE CONCRETE 10X20 GRADE 300 (DURABILITY LEVEL B22.5)

* STEEL < 10 .Rs = 2250 kg/cm2 (AI STEEL)

* STEEL >= 10 .Rs = 2800 kg/cm2 (AII STEEL)

NOTES :


2


FOUNDATION PLAN

TL:1/25



Ho Chi Minh City College of Construction

300

SECTION P2


8300


CONNECTION SEE DETAILS


16.6M PILE DIAGRAM


SECTION P1


Foundation design guide

8300


200


Figure 3.22 Pile design drawing

250


2500


2050 (Þ6/50)


8300


3300 (Þ6/150)


8

DETAILS OF PILE SECTION P2


2050 (Þ6/50)


2500


250


200


2x2ị12


A


HOLE20


125 125

300

2Þ1 4 1 MeC ị16 (SMOOTH) 7


4 2x2ị12A


3

5 STEEL MESH Þ6@50


4Þ18


DETAILS OF PILE SECTION P1


20 360 20

9

L 63x63x6 L=360


15

DIG A 50x50 HOLE ON 1 SIDE OF THE PILE TO COMPACT THE CONCRETE AT THE HEAD OF THE PILE


9

L 63x63x6 L=360


-190x8


-190x85

8

254

270

254


10 -280x10

280


FACTORY WELDING



5 5

-190x8


WELDING STEEL MESH TO LONGITUDINAL STEEL OF PILES


4Þ18 5 STEEL MESH 3


5 -190x6

254


8

25

4Þ124


300

5 2

250

2 5

8

WELDING JOINT EDGE Hh=8


250

300

60 50

5 254


5

4 2

-190x8 254

WELDING JOINTSANGLE STEEL

254 5 1 8

Þ6@50

1 8 4Þ18

WELDING TO PLATE

-190x208x6

1 4Þ18


200

110 50 110


15 270

300

8


200

254 8

270

L 63x63x6 9

STEEL MESH WELDING DETAILSINTO STEEL CHU Û

8 254

270

8 JOINT EDGE WELDING


Chapter 3: Pile foundation design

8 254

270

25

SPIRAL BELT 62

@ 50-150

DETAILS 1

DETAILS OF THE PASSAGE OF NO CATHOLIC 1-1


11 2Þ1 4

11 2Þ1 4


300

420

125 50 125

AA CAFEBB CUT

- PILE 300x300, L=16.6M

- STONE CONCRETE 10x20; GRADE 300 (DURABILITY LEVEL B22.5)

- LOW BONE: ị<10 Rs=2250KG/CM2. (AI)

ị >=10 Rs=2800KG/CM2. (AII)

- DESIGN LOAD Ptk = 70.5T

- TEST PILES ACCORDING TO TCVN 9393:2012,

2-CYCLE TEST, MAXIMUM TEST LOAD LEVEL 2Ptk.

- WELD STRENGTH Rgh>=1700 KG/CM2.

- WELDING ENTIRE CONTACT LENGTH, Hh>=6MM.

NOTES​

10

420

20


10

-270x10

Page | 103

4 2x2ị12 270

5 STEEL MESH

360

20

9

L 63x63x6 L=360

Hole 20 x 18, L=590

12

30 210 30

210

30

30

30

210

-190x85

-270x10 270

10 -270x10

30 210 30

125 50 125

300

270

HOLE20

218

218

50

30

LOW GRID 6a50

1Þ20

LOOK CC

3 Þ6@50

200

200

254

PILE CAP DETAILS

CUT 2-2PILE CAP DETAILS

(PILE CONNECTION)




INDEX

MISSION 2

IMPLEMENTATION OF FOUNDATION PROJECT 2

Code 4

SINGLE FOUNDATION DESIGN SECTION 4

CHAPTER 1 14

GENERAL INSTRUCTIONS 14

1.1 IMPLEMENTATION CONTENT 14

1.2 CALCULATION DESCRIPTION REQUIREMENTS 14

1.3 DRAWING REQUIREMENTS 17

CHAPTER 1 DESIGN OF SHALL FOUNDATIONS 18

2.1 SINGLE FOUNDATION CALCULATION DESIGN PROCEDURE 18

2.1.1Determine the load on the foundation 18

2.1.2 Assessment of hydrogeological conditions 18

2.1.3Determine foundation depth 18

2.1.4 Determining the preliminary dimensions of the foundation 19

2.1.5 Checking the stress under the foundation 22

2.1.6 Calculation and settlement testing 23

2.1.7 Determine the working height of the foundation (Hm) 29

2.1.8 Calculation of reinforcement 32

2.2 SINGLE FOUNDATION DESIGN CALCULATION EXAMPLE 34

2.2.1Determine the load on the foundation 35

2.2.2 Summary of hydrogeological conditions 35

2.2.3 Choosing the depth of foundation 35

2.2.4 Determine the preliminary size of the foundation 36

2.2.5 Checking stress under the foundation 37

2.2.6 Calculation and settlement testing 38

2.2.7 Determine the working height of the foundation 40

2.2.8 Calculation of reinforcement 42

2.2.9 Drawing representation 45

Chapter 3 PILE FOUNDATION DESIGN 46

3.1 DESIGN PROCEDURE FOR LOW PILE FOUNDATION 46

3.1.1Determine the load on the foundation 46

3.1.2 Assessment of terrain, geology and hydrology conditions 46

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